The determinant $\left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \\ -6 & 9 + x^2 & 3 \\ -2 & 3 & 1 + x^2 \end{array} \right|$ for $x \neq 0$ is not divisible by:

If $A = \begin{bmatrix} 2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15 \end{bmatrix}$,then the rank of $A$ is equal to

Let for $i = 1, 2, 3$,$p_i(x)$ be a polynomial of degree $2$ in $x$,$p'_i(x)$ and $p''_i(x)$ be the first and second order derivatives of $p_i(x)$ respectively. Let $A(x) = \begin{bmatrix} p_1(x) & p'_1(x) & p''_1(x) \\ p_2(x) & p'_2(x) & p''_2(x) \\ p_3(x) & p'_3(x) & p''_3(x) \end{bmatrix}$ and $B(x) = [A(x)]^T A(x)$. Then the determinant of $B(x)$

The rank of the following matrix $A$ is
$A = \begin{bmatrix} 1 & -2 & 3 & -4 \\ 2 & 9 & 4 & 5 \\ 4 & 5 & 10 & -3 \\ 1 & 11 & -1 & 9 \end{bmatrix}$

Let $A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & k-1 \\ 0 & 0 & k-1 & 1 \end{bmatrix}$ and $k \in R$. Then,the value of $k$,if it exists,for which the rank of $A$ is $2$,is

If the rank of the matrix $A=\begin{bmatrix} 1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k \end{bmatrix}$ is $2$ and $k$ is a real number,then $k$ is a root of the following quadratic equation: